Sudden Perturbation Approximation Question

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In the context of beta decay from H3 to He3+, the sudden perturbation approximation is applied to determine the transition probability of an electron from the 1s state of H3 to the |n=16, l=3, m=0> state of He3+. The calculated probability is zero due to the orthonormality of spherical harmonics, which indicates that the inner product is zero unless the quantum numbers l and m match. Since the initial state has no angular momentum (l=0), it cannot transition to a state with l=3, confirming that the probability is indeed zero. This result highlights that a particle in an s state cannot possess angular momentum, aligning with the sudden approximation's assumption of unchanged orbital angular momentum during rapid transitions. The discussion clarifies the reasoning behind the zero probability outcome.
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Homework Statement


In a beta decay H3 -> He3+, use the sudden perturbation approximation to determine the probability of that an electron initially in the 1s state of H3 will end up in the |n=16,l=3,m=0> state of He3+


Homework Equations


|<n'l'm'|nlm>|^2


The Attempt at a Solution



I actually know the answer to this but I am not clear as to why and I am wondering if there is an easier way to determine the solution.

The answer comes out to be 0. When integrating the wavefunctions |n'l'm'> and |nlm> in spherical coordinates to calculate the inner product (i.e. the probability amplitude), the integral over d\theta returns 0. Is there an easier way to see this other than going through the calculations?

How is this result interpreted?

Thanks,

jsc
 
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jsc314159 said:

Homework Statement


In a beta decay H3 -> He3+, use the sudden perturbation approximation to determine the probability of that an electron initially in the 1s state of H3 will end up in the |n=16,l=3,m=0> state of He3+


Homework Equations


|<n'l'm'|nlm>|^2


The Attempt at a Solution



I actually know the answer to this but I am not clear as to why and I am wondering if there is an easier way to determine the solution.

The answer comes out to be 0. When integrating the wavefunctions |n'l'm'> and |nlm> in spherical coordinates to calculate the inner product (i.e. the probability amplitude), the integral over d\theta returns 0. Is there an easier way to see this other than going through the calculations?

How is this result interpreted?

Thanks,

jsc

The spherical harmonics are orthonormal so <l' m' | l m> is zero unless l=l' and m= m'. Clearly here the result is zero since |3,0> is orthogonal to |0,0>

Physically, it simply says that a particle in an s state has no angular momentum so the probability of it being observed with l=3 is zero. The sudden approximation simply assumes that the transition was so quick that the orbital angular momentum of the electron remained unchanged.
 
Last edited:
Thanks nrged.

That makes it clear.
 

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